Question

The equation $(1-tan\theta)(1+tan\theta)sec^2\theta + 2^{tan^2\theta}=0$ has:

• A. No solution in the interval $(-\frac{\pi}{2},0)$
• B. Two solutions in the interval $(-\frac{\pi}{2},0)$
• C. No solution in the interval $(0,\frac{\pi}{2})$
• D. Two solutions in the interval $(0,\frac{\pi}{2})$

Answer 1

Answer: (B), (D)

B, D

Answer 2

The given equation is: $(1-\tan\theta)(1+\tan\theta)\sec^2\theta + 2^{\tan^2\theta}=0$

First, simplify the terms:

1. $(1-\tan\theta)(1+\tan\theta) = 1^2 - \tan^2\theta = 1 - \tan^2\theta$.
2. We know that $\sec^2\theta = 1 + \tan^2\theta$.

Substitute these into the equation: $(1 - \tan^2\theta)(1 + \tan^2\theta) + 2^{\tan^2\theta} = 0$

Again, apply the difference of squares formula: $1^2 - (\tan^2\theta)^2 + 2^{\tan^2\theta} = 0$ $1 - \tan^4\theta + 2^{\tan^2\theta} = 0$

Let $x = \tan^2\theta$. Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\tan\theta$ can take any real value. Therefore, $\tan^2\theta = x$ must be non-negative, i.e., $x \ge 0$.

The equation becomes: $1 - x^2 + 2^x = 0$ Rearrange the terms: $2^x = x^2 - 1$

To find the solutions for $x$, we can analyze the function $h(x) = 2^x - x^2 + 1$. We are looking for $x \ge 0$ such that $h(x)=0$.

Let's evaluate $h(x)$ for some non-negative integer values:

• For $x=0$: $h(0) = 2^0 - 0^2 + 1 = 1 - 0 + 1 = 2$.
• For $x=1$: $h(1) = 2^1 - 1^2 + 1 = 2 - 1 + 1 = 2$.
• For $x=2$: $h(2) = 2^2 - 2^2 + 1 = 4 - 4 + 1 = 1$.
• For $x=3$: $h(3) = 2^3 - 3^2 + 1 = 8 - 9 + 1 = 0$.

So, $x=3$ is a solution.

Let's check for other solutions. Consider the derivative of $h(x)$: $h'(x) = \frac{d}{dx}(2^x - x^2 + 1) = 2^x \ln 2 - 2x$.

Evaluate $h'(x)$ at $x=3$: $h'(3) = 2^3 \ln 2 - 2(3) = 8 \ln 2 - 6$. Using $\ln 2 \approx 0.693$: $h'(3) \approx 8 \times 0.693 - 6 = 5.544 - 6 = -0.456$. Since $h'(3) < 0$, the function $h(x)$ is decreasing at $x=3$. This means for $x$ slightly greater than $3$, $h(x)$ will become negative.

Let's check $h(4)$: $h(4) = 2^4 - 4^2 + 1 = 16 - 16 + 1 = 1$. Since $h(3)=0$ and $h(4)=1$, and $h(x)$ is continuous, and $h'(3)<0$ implies $h(x)$ goes negative after $x=3$, there must be another root between $3$ and $4$. To confirm $h(x)$ becomes negative, let's check $h(3.2)$: $h(3.2) = 2^{3.2} - (3.2)^2 + 1 \approx 9.189 - 10.24 + 1 = -0.051$. Since $h(3.2) < 0$ and $h(4) = 1$, by the Intermediate Value Theorem, there exists another root $x_0 \in (3.2, 4)$.

So, we have two distinct solutions for $x = \tan^2\theta$:

1. $x = 3$
2. $x = x_0$, where $x_0 \in (3.2, 4)$

Now, we convert these solutions back to $\theta$:

Case 1: $\tan^2\theta = 3$ $\tan\theta = \pm\sqrt{3}$

• If $\tan\theta = \sqrt{3}$, then $\theta = \frac{\pi}{3}$. This solution lies in the interval $(0, \frac{\pi}{2})$.
• If $\tan\theta = -\sqrt{3}$, then $\theta = -\frac{\pi}{3}$. This solution lies in the interval $(-\frac{\pi}{2}, 0)$.

Case 2: $\tan^2\theta = x_0$, where $x_0 \in (3.2, 4)$ $\tan\theta = \pm\sqrt{x_0}$ Since $x_0 \in (3.2, 4)$, we have $\sqrt{x_0} \in (\sqrt{3.2}, \sqrt{4}) = (\sqrt{3.2}, 2)$. Note that $\sqrt{3.2} \approx 1.788$.

• If $\tan\theta = \sqrt{x_0}$: Since $\sqrt{x_0} > 0$, $\theta = \arctan(\sqrt{x_0})$. This solution lies in $(0, \frac{\pi}{2})$. Since $\sqrt{x_0} \ne \sqrt{3}$, this solution is distinct from $\frac{\pi}{3}$.
• If $\tan\theta = -\sqrt{x_0}$: Since $-\sqrt{x_0} < 0$, $\theta = \arctan(-\sqrt{x_0}) = -\arctan(\sqrt{x_0})$. This solution lies in $(-\frac{\pi}{2}, 0)$. Since $-\sqrt{x_0} \ne -\sqrt{3}$, this solution is distinct from $-\frac{\pi}{3}$.

In summary:

• In the interval $(-\frac{\pi}{2}, 0)$, we have two distinct solutions: $\theta = -\frac{\pi}{3}$ and $\theta = -\arctan(\sqrt{x_0})$.
• In the interval $(0, \frac{\pi}{2})$, we have two distinct solutions: $\theta = \frac{\pi}{3}$ and $\theta = \arctan(\sqrt{x_0})$.

Therefore, both options (B) and (D) are correct.